Problem 77 - Introductory and Intermediate Algebra for College Students 4th

Question

A new car worth 45,000 dollars is depreciating in value by 5000 dollars per year. The mathematical model $$y=-5000 x+45,000$$ describes the car's value, \(y,\) in dollars, after \(x\) years. a. Find the \(x\) -intercept. Describe what this means in terms of the car's value. b. Find the \(y\) -intercept. Describe what this means in terms of the car's value. c. Use the intercepts to graph the linear equation. Because \(x\) and \(y\) must be nonnegative (why?), limit your graph to quadrant I and its boundaries. d. Use your graph to estimate the car's value after five years.

Step-by-Step Solution

Answer

a. The x-intercept is 9 years. It means that the car will have no value after 9 years. \n b. The y-intercept is $45,000. It means that the car was worth $45,000 initially. \n c. the graph of the equation is a straight line, starting at (0, 45000) and going through point (9, 0). \n d. The car is estimated to be worth about $25,000 after five years.

Step 1: Find the x-intercept

To find the x-intercept, set y = 0 and solve for x in the equation \(y = -5000x + 45000\). Doing so, we obtain \(x = \frac{45000}{5000} = 9\) years.

Step 2: Interpret the x-intercept

The x-intercept represents the number of years after which the car's value depreciates to zero. In this case, it takes 9 years for the car to have no resale value.

Step 3: Find the y-intercept

To find the y-intercept, set x = 0 and solve for y in the equation \(y = -5000x + 45000\). We obtain \(y = 45000\) dollars.

Step 4: Interpret the y-intercept

The y-intercept typically represents the initial value of the dependent variable. Here, it represents the initial cost of the car, which is $45,000.

Step 5: Graph the equation

To graph the equation, plot the y-intercept (0, 45000) and the x-intercept (9, 0). Draw a straight line connecting these points. Limit the graph to quadrant I as both x and y are nonnegative.

Step 6: Estimate the value after 5 years

From the constructed graph, estimate the value of the car after 5 years. Since the graph might not be to scale, it's an approximation. For the exact value, substitute \(x=5\) in the equation to get \(y= -5000(5) +45000 = \$25000\).

Key Concepts

Intercepts of a Linear Equation

Understanding intercepts of a linear equation is crucial for graphing and interpreting the behavior of the line. An x-intercept is a point where the graph crosses the x-axis, which means the value of y is zero at that point. To find it, simply set the y-value to zero and solve for x. The x-intercept in our example was found to be 9 years. This number is significant in the context of the linear depreciation model, as it indicates the time it takes for the car to have no monetary value.

Similarly, the y-intercept is the point where the graph crosses the y-axis, indicating the x-value is zero. For the given equation, when x is set to zero, the y-value was calculated as $45,000. This figure represents the initial value of the car before any depreciation occurs. Collectively, intercepts provide a straightforward method to sketch the behavior of linear relationships over time. They are the two essential points you need to graph a straight line - the essence of a linear model.

Graphing Linear Equations

Graphing linear equations involves plotting points and drawing lines representing the equation's solution set. The process is foundational in visualizing how variables in the equation relate to each other. Begin by identifying intercepts, as these anchor points simplify the creation of an accurate graph. In our car depreciation example, plotting the points (0, 45000) and (9, 0) sets the stage. A straight line is then drawn between them to depict the car's value as it decreases over time.

Why do we limit the graph to the first quadrant and its boundaries? Since the car's value and the number of years cannot be negative, representing them on other quadrants would make no sense in this context. Graphing not only aids in visual understanding but also enables estimations; after graphically depicting the car's value decrease, one can approximate its value at any point along the timeline within the graph, such as the car's value after five years.

Linear Equation Application

The application of linear equations can be found in various real-world scenarios, with the linear depreciation model being a prime example. Such equations allow us to model and predict values over time, like the value of a car as it depreciates annually. This particular linear equation, represented by \(y = -5000x + 45000\), not only explains the value change per annum but also forecasts the future worth of the vehicle.

Applications extend beyond economics into sciences, engineering, and daily life. They serve as tools for making informed decisions based on data and trends. In the context of our example, knowing that the car will depreciate to zero value over nine years helps a potential buyer understand the investment's lifespan. The simplicity and predictability of linear models make them quintessential for beginner-level understanding of algebra's utility in solving practical problems.